Question

If the roots of $(p-q)x^2+(q-r)x+(r-p)=0$ are real and equal, then which of the following is/are correct?

• A. $r,p,q$ are in A.P.
• B. $p,q,r$ are in A.P.
• C. Product of the roots is $1$
• D. $q,r,p$ are in A.P.

Answer 1

Answer: (A), (C)

The correct options are
A $r,p,q$ are in A.P.
C Product of the roots is $1$

$(p-q)x^2+(q-r)x+(r-p)=0$
By observation, $x=1$ satisfied the above equation,
As the given equation is real and equal, so both roots are $1,1$
Product of roots
$\frac{r-p}{p-q}=1\Rightarrow r-p=p-q\Rightarrow 2p=q+r$
Therefore,
$q,p,r$ or $r,p,q$ are in an A.P.